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miss Akunina [59]
3 years ago
5

Which statement best reflects the solution(s) of the equation?

Mathematics
1 answer:
Inessa [10]3 years ago
5 0

x=2 is only solution while x=1 is extraneous solution

Option C is correct.

Step-by-step explanation:

We need to solve the equation \frac{1}{x-1}+\frac{2}{x}=\frac{x}{x-1} and find values of x.

Solving:

Find the LCM of denominators x-1,x and x-1. The LCM is x(x-1)

Multiply the entire equation with x(x-1)

\frac{1}{x-1}+\frac{2}{x}=\frac{x}{x-1}\\\frac{1}{x-1}*x(x-1)+\frac{2}{x}*x(x-1)=\frac{x}{x-1}*x(x-1)\\Cancelling\,\,out\,\,the\,\,same\,\,terms:\\x+2(x-1)=x^2\\x+2x-2=x^2\\3x-2=x^2\\x^2-3x+2=0

Now, factoring the term:

x^2-2x-x+2=0\\x(x-2)-1(x-2)=0\\(x-1)(x-2)=0\\x-1=0\,\,and\,\, x-2=0\\x=1\,\,and\,\, x=2

The values of x are x=1 and x=2

Checking for extraneous roots:

Extraneous roots: The root that is the solution of the equation but when we put it in the equation the answer turns out not to be right.

If we put x=1 in the equation, \frac{1}{x-1}+\frac{2}{x}=\frac{x}{x-1}  the denominator becomes zero i.e

\frac{1}{1-1}+\frac{2}{1}=\frac{1}{1-1}\\\frac{1}{0}+2=\frac{1}{0}

which is not correct as in fraction anything divided by zero is undefined. So, x=1 is an extraneous solution.

If we put x=2  in the equation,

\frac{1}{x-1}+\frac{2}{x}=\frac{x}{x-1}

\frac{1}{2-1}+\frac{2}{2}=\frac{2}{2-1}\\\frac{1}{1}+1=\frac{2}{1}\\1+1=2\\2=2

So, x=2 is only solution while x=1 is extraneous solution

Option C is correct.

Keywords: Solving Equations and checking extraneous solution

Learn more about Solving Equations and checking extraneous solution at:

  • brainly.com/question/1626495
  • brainly.com/question/2959656
  • brainly.com/question/2456302

#learnwithBrainly

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